 Solution

For odd $p$, ${\mathbb{Q}}_p$ contains exactly the $p-1$ Teichmuller roots of unity; for $p=2$ it contains $\{\pm 1\}$, so there are two. For odd $p$, adjoining ${\zeta }_p$ adds the $p$ roots of unity of $p$-power order and no primitive $p^2$th root, because the latter would enlarge the degree by $p$. Combining the coprime-order groups givesFor $p=2$, ${\zeta }_2=-1$ already lies in ${\mathbb{Q}}_2$, so the answer remains two.

Solved by gpt-5.6-sol high.